Preimage of the complement

Question

For a subset $X$ of $Y$, $f^{-1}(X') = (f^{-1}(X))'$. Here $'$ denotes the complementary set.

Is this true? If so, what would be a proof for it? If it isn't, what would be a counterexample? I'm completely lost.

Maybe $X'$ is the complementary set of $X$? – Emanuele Paolini Nov 17 '14 at 21:54 — Emanuele Paolini, Nov 17 '14 at 21:54

egreg · Accepted Answer · 2018-02-07T17:33:02.203

18

You apparently are given a function $f\colon A\to Y$. Let's try.

An element $a\in A$ belongs to $f^{-1}(X')$ if and only if $f(a)\in X'$ or $f(a)\notin X$, which means $a\notin f^{-1}(X)$ or $a\in (f^{-1}(X))'$.

I like to set these chains in a column $$ a\in f^{-1}(X')\\ f(a)\in X'\\ f(a)\notin X\\ a\notin f^{-1}(X)\\ a\in (f^{-1}(X))' $$ and then I observe that any line is a consequence of the line above, but also of the line below.

edited Feb 07 '18 at 17:33

answered Nov 17 '14 at 22:01

egreg

244,946

2

In summary, $f^{-1}(Y\setminus X)=f^{-1}(Y)\setminus f^{-1}(X)$, right? – Filippo Jun 06 '22 at 10:40
2

@Filippo Yes, that's correct. – egreg Jun 06 '22 at 10:44

Preimage of the complement

1 Answers1

Linked